If $\mathrm{\theta }$ is the angle between unit vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}},$ then $\frac{\left(1-\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right)}{\left(1+\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}\right)}$ is equal to

Given,

$\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}},$are unit vectors

$\overrightarrow{A}.\overrightarrow{B}=\left|1\right|\left|1\right|\cos \theta$

$\frac{1-\overrightarrow{A}.\overrightarrow{B}}{1+\overrightarrow{A}.\overrightarrow{B}}=\frac{1-\cos \theta }{1+\cos \theta }=\frac{2{\sin }^2{\displaystyle \frac{\theta }{2}}}{2{\cos }^2{\displaystyle \frac{\theta }{2}}}={\tan }^2\frac{\theta }{2}$

Hence the correct answer is ${\tan }^2\frac{\theta }{2}.$